Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
og 12 question 11 in ps
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ha just answered my own question, realized that the real question to ask is "what raised to the third power equals 10 ^ -6?" and of course, 10^ -2 is the answer because when numbers with the same base are multiplied by one another, their exponents are simply added. yay me
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Already discussed,aerodan1 wrote:Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
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Third root means cube root, and cube root of x is given by x^(1/3). When x is 10^-6, its cube root is given by (10^-6)^(1/3) and since the laws of indices say that (a^m)^n = (a)^(m n), therefore (10^-6)^(1/3) = (10)^-6(1/3) = (10)^-2.aerodan1 wrote:Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
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Sanjeev K Saxena
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